A) 80%
B) 90%
C) 87.5%
D) 82.5%
Correct Answer: C
Solution :
The power of the battery, when charged is given by \[M=\frac{1000\times {{k}_{b}}\times w}{W\times \Delta {{T}_{b}}}\] The electrical energy dissipated is given by \[\Delta {{T}_{b}}=\frac{1000\times {{k}_{b}}\times 10}{100\times 100}\] \[{{E}_{1}}={{V}_{1}}{{l}_{1}}{{t}_{1}}=15\times 10\times 8\] \[=1200Wh\] Similarly, the electrical energy dissipated during the discharge of battery is given by \[\left( {{\text{R}}_{\text{1}}}\text{-X and }{{\text{R}}_{\text{2}}}\text{-X} \right)\] The electrical energy dissipated is given by \[{{E}_{2}}={{V}_{2}}{{l}_{2}}{{t}_{2}}\] \[=14\times 5\times 15\] \[=1050Wh\] Similarly, the electrical energy dissipated during the discharge of battery is given by \[\begin{align} & OO \\ & |||| \\ & C{{H}_{3}}-C-C{{H}_{2}}-C-O{{C}_{2}}{{H}_{5}}\rightleftharpoons \\ & \left( keto \right) \\ \end{align}\] \[\begin{align} & OHO \\ & ||| \\ & C{{H}_{3}}-C=CH-C-O{{C}_{2}}{{H}_{5}} \\ & \left( enol \right) \\ \end{align}\] = 1050 Wh Hence, watt-hour efficiency of the battery is given by \[N{{a}_{2}}S\] \[N{{a}_{2}}S+{{\left( C{{H}_{3}}COO \right)}_{2}}Pb\to \]
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